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Solve-the-system-x-C-y-1-20-x-1-C-y-10-




Question Number 57074 by Tawa1 last updated on 30/Mar/19
Solve the system.         ^x C_(y + 1)   =  20,            ^(x − 1) C_y   =  10
Solvethesystem.xCy+1=20,x1Cy=10
Answered by mr W last updated on 30/Mar/19
  ^(x − 1) C_y   =  10  ⇒(((x−1)!)/((x−1−y)!y!))=10      ^x C_(y + 1)   =  20  ⇒((x!)/((x−y−1)!(y+1)!))=20  ⇒((x(x−1)!)/((x−1−y)!(y+1)y!))=20  ⇒(x/((y+1)))×10=20  ⇒(x/(y+1))=2  ⇒x=2y+2    ⇒(((2y+1)!)/((y+1)!y!))=10  ⇒(((y+y+1)!)/((y+1)!y!))=10  ⇒(((y+1+1)(y+1+2)...(y+1+y))/(1×2×...y))=10  =(((y+1)/1)+1)(((y+1)/2)+1)...(((y+1)/y)+1)=10  since each term is >2, we may have  at most 3 terms to get product 10,  i.e. y≤3.  with y=3: ⇒x=8   ^(x − 1) C_y   = ^7 C_3 =35≠10  with y=2: ⇒x=6   ^(x − 1) C_y   = ^5 C_2 =10 ⇒ok    ⇒solution is x=6, y=2
x1Cy=10(x1)!(x1y)!y!=10xCy+1=20x!(xy1)!(y+1)!=20x(x1)!(x1y)!(y+1)y!=20x(y+1)×10=20xy+1=2x=2y+2(2y+1)!(y+1)!y!=10(y+y+1)!(y+1)!y!=10(y+1+1)(y+1+2)(y+1+y)1×2×y=10=(y+11+1)(y+12+1)(y+1y+1)=10sinceeachtermis>2,wemayhaveatmost3termstogetproduct10,i.e.y3.withy=3:x=8x1Cy=7C3=3510withy=2:x=6x1Cy=5C2=10oksolutionisx=6,y=2
Commented by Tawa1 last updated on 30/Mar/19
God bless you sir. I appreciate.
Godblessyousir.Iappreciate.
Commented by peter frank last updated on 30/Mar/19
nice work
nicework

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